A-Level Further Mathematics: Differential Equations

Differential equations form the mathematical backbone for modeling change across physics, engineering, economics, and biology. Mastering them allows you to move from describing static relationships to predicting dynamic systems—from the swing of a pendulum to the growth of a population.

First Order Differential Equations: Foundational Techniques

A first order differential equation involves a function, $y$, its first derivative, $\frac{dy}{dx}$, and the variable $x$. The general form is $\frac{dy}{dx}=f(x,y)$. The order is defined by the highest derivative present. Our first goal is to find the general solution, which contains an arbitrary constant of integration. For many problems, you will then use an initial condition (a known pair of $x$ and $y$ values) to find a particular solution.

The simplest method is separation of variables. This applies when you can manipulate the equation to have all terms involving $y$ on one side (with $dy$) and all terms involving $x$ on the other (with $dx$). The process is straightforward: 1) Separate the variables, 2) Integrate both sides, 3) Include the constant of integration, $C$, and 4) Simplify if possible.

Example: Solve $\frac{dy}{dx}=2xy$, given $y=1$ when $x=0$.

1. Separate: $\frac{1}{y}dy=2xdx$.
2. Integrate: $\int \frac{1}{y}dy=\int 2xdx$ → $\ln |y|=x^2+C$.
3. General solution: $|y|=e^{x^2+C}=e^C{e}^{x^2}$. Let $A=e^C$, giving $y=A{e}^{x^2}$.
4. Apply initial condition ($x=0,y=1$): $1=A{e}^0$ → $A=1$.

Particular solution: $y=e^{x^2}$.

When variables cannot be separated, the integrating factor method is used for linear first order equations of the form $\frac{dy}{dx}+P(x)y=Q(x)$. The strategy is to multiply the entire equation by a cleverly chosen function, the integrating factor (IF), which makes the left-hand side a perfect derivative. The integrating factor is defined as $IF=e^{\int P(x)dx}$.

Example: Solve $x\frac{dy}{dx}-2y=x^3$.

1. Rewrite in standard form: $\frac{dy}{dx}-\frac{2}{x}y=x^2$. Here, $P(x)=-\frac{2}{x}$, $Q(x)=x^2$.
2. Find IF: $\int P(x)dx=\int -\frac{2}{x}dx=-2\ln |x|=\ln (x^{-2})$. Therefore, $IF=e^{\ln (x^{-2})}=x^{-2}=\frac{1}{x^2}$.
3. Multiply through by IF: $\frac{1}{x^2}\frac{dy}{dx}-\frac{2}{x^3}y=1$.

The left side is now the derivative of $\left(y\cdot \frac{1}{x^2}\right)$.

1. Integrate both sides with respect to $x$: $\frac{d}{dx}\left(\frac{y}{x^2}\right)=1$ → $\frac{y}{x^2}=\int 1dx=x+C$.
2. General solution: $y=x^3+C{x}^2$.

Second Order Linear Equations with Constant Coefficients

These equations have the standard form $a\frac{d^{2}y}{d{x}^2}+b\frac{dy}{dx}+cy=f(x)$, where $a$, $b$, and $c$ are constants. The solution, $y(x)$, is the sum of two parts: the complementary function (CF) and the particular integral (PI). That is, $y=CF+PI$.

The complementary function is the general solution to the homogeneous equation (where $f(x)=0$). To find it, we solve the auxiliary equation: $a{m}^2+bm+c=0$. The roots $m_1$ and $m_2$ of this quadratic determine the form of the CF:

• Real, distinct roots ($m_1\ne m_2$): $CF=A{e}^{m_{1}x}+B{e}^{m_{2}x}$.
• Real, equal roots ($m_1=m_2=m$): $CF=(A+Bx)e^{mx}$.
• Complex roots ($m=p\pm iq$): $CF=e^{px}(A\cos qx+B\sin qx)$.

Here, $A$ and $B$ are arbitrary constants to be determined later by boundary conditions (values given at different $x$ points) or initial conditions (values given at a single point, often $x=0$).

Finding the Particular Integral

The particular integral is any solution that satisfies the full, non-homogeneous equation. Its form is guessed based on the function $f(x)$. This "trial function" method is systematic:

• If $f(x)=k$ (a constant), try $PI=\lambda$, a constant.
• If $f(x)=px+q$ (a polynomial), try a general polynomial of the same degree: $PI=\lambda x+\mu$.
• If $f(x)=e^{kx}$, try $PI=\lambda e^{kx}$.
• If $f(x)=\sin kx$ or $\cos kx$, try $PI=\lambda \sin kx+\mu \cos kx$.

Crucial Check: If your trial function is already part of the Complementary Function, you must multiply it by $x$ (or $x^2$ if it's a repeated root case). Substitute the trial PI and its derivatives into the original ODE, and equate coefficients to find the constants like $\lambda$ and $\mu$.

Example: Solve $\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+2y=2e^{3x}$.

1. Auxiliary equation: $m^2-3m+2=0$ → $(m-1)(m-2)=0$ → $m=1,2$.

CF: $y_c=A{e}^x+B{e}^{2x}$.

1. For $f(x)=2e^{3x}$, try $PI=\lambda e^{3x}$.

Then $\frac{d}{dx}(PI)=3\lambda e^{3x}$ and $\frac{d^2}{d{x}^2}(PI)=9\lambda e^{3x}$.

1. Substitute into ODE: $(9\lambda e^{3x})-3(3\lambda e^{3x})+2(\lambda e^{3x})=2e^{3x}$.

This simplifies to $2\lambda e^{3x}=2e^{3x}$ → $\lambda =1$. PI: $y_p=e^{3x}$.

1. General solution: $y=A{e}^x+B{e}^{2x}+e^{3x}$.

Modelling Real-World Systems

Differential equations are powerful because they translate physical laws into solvable mathematics.

• Damped Mechanical Oscillations: A mass on a spring with a damper is modeled by $m\ddot{x}+c\dot{x}+kx=0$, a second-order homogeneous ODE. The roots of the auxiliary equation determine if the system is over-damped (real distinct roots), critically damped (real equal roots), or under-damped (complex roots leading to oscillatory decay).
• Population Dynamics: The logistic growth model, $\frac{dP}{dt}=kP(M-P)$, is a first-order, separable ODE that refines exponential growth by including a carrying capacity, $M$.
• Electrical Circuits: An RLC circuit (Resistor, Inductor, Capacitor) is governed by $L\frac{d^{2}Q}{d{t}^2}+R\frac{dQ}{dt}+\frac{1}{C}Q=E(t)$, directly analogous to the damped oscillation equation, with charge $Q$ corresponding to displacement.

In all these applications, you must use the given initial conditions (like initial position and velocity, or initial population) or boundary conditions to find the specific constants $A$ and $B$ in your final solution, moving from a general description of possible behaviors to the precise trajectory of your specific system.

Common Pitfalls

1. Misapplying the Integrating Factor: The most common error is failing to write the first-order linear equation in the standard form $\frac{dy}{dx}+P(x)y=Q(x)$ before calculating $IF=e^{\int P(x)dx}$. If the coefficient of $\frac{dy}{dx}$ is not 1, you must divide the entire equation by it first.
2. Incorrect PI Guess When It's Part of the CF: If your trial function for the PI appears in the Complementary Function, your initial substitution will yield $0=f(x)$, which is impossible. Always check and multiply by $x$. For example, if solving $\ddot{x}-x=e^t$ and your CF is $A{e}^t+B{e}^{-t}$, your first guess for the PI for $e^t$ would be $\lambda e^t$. Since $e^t$ is in the CF, you must instead try $PI=\lambda t{e}^t$.
3. Forgetting the Arbitrary Constants on the CF: The general solution is $y=CF+PI$. The constants $A$ and $B$ belong only to the CF. A frequent mistake is to add them to the PI or to omit them entirely before applying boundary conditions. They remain undetermined until the very last step.
4. Poor Algebraic Manipulation with Integration: When separating variables or integrating, careful algebra is paramount. A sign error or misplacement of a constant early on will propagate through the entire solution. Always check your integration and exponentiation steps, especially when going from $\ln |y|$ to $y$.

Summary

• First-order ODEs are tackled via separation of variables or the integrating factor method for linear equations. The solution incorporates a constant found using an initial condition.
• The general solution to a second-order linear ODE with constant coefficients is $y=CF+PI$. You find the Complementary Function (CF) by solving the auxiliary equation, which yields forms based on real or complex roots.
• The Particular Integral (PI) is found using a trial function based on $f(x)$. You must adjust the trial (by multiplying by $x$) if it duplicates part of the CF.
• These techniques model dynamic systems like damped oscillations, population growth, and electrical circuits. Final, specific solutions require applying boundary or initial conditions to determine the arbitrary constants in the CF.